A capacitor has a peak current of 310 μa when the peak voltage at 300 khz is 3.0 v.

I think the question for this problem, is to find the capacitance. The working equation would be:

C = Q/V,

where

Q = It

Since f = 1/t, t = (1/300 kHz) (1 kHz/1,000 Hz) = 3.33*10⁻⁶ s

So,

Q = (310*10⁻⁶ A) (3.33*10⁻⁶ s)

Q = 1.033*10⁻⁹ C

Thus,

C = (1.033*10⁻⁹ C) / (3 V)

C = 3.44*10⁻¹⁰ F